# Why are units called units?

Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?

• I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide. – Omnomnomnom Oct 5 '15 at 5:54
• I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers) – Eric Wofsey Oct 5 '15 at 5:58

"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.

In the Sciences a unit is a fixed amount of a quantity in question. Every other amount is a multiple of the fixed amount. In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.

• Is this true about abstract algebra too? – Arman Malekzadeh Jul 26 '17 at 19:30

To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.

In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $$u$$ is a unit of a ring $$R$$, then there is a $$v\in R$$ such that $$uv=vu=1$$. Thus, for any $$x\in R$$, one can rewrite $$x$$ "in terms of the unit $$u$$" like so: $$x=x(vu)=(xv)u$$. Thus, $$x$$ is a multiple of $$u$$.

This can be understood well in seeing why a ring $$F$$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are $$\{0\}$$ and $$F$$.

Note every element in a field is a unit $$u$$. It follows that if we rewrite any element $$x\in F$$ in terms of that unit, we will get $$x=ku$$ for some $$k\in F$$. Now consider that any ideal is a kernel of a ring homomorphism $$f:F\to R$$. $$f$$'s kernel could be $$\{0\}$$, in which case $$f$$ is injective, and the ideal it corresponds to is $$\{0\}$$.

Otherwise, some element (and thus unit) $$u$$ maps to zero. Since we can write any $$x$$ as $$x=ku$$, that means $$f(x)=f(k)f(u)=0.$$ So the kernel is all of $$F$$.

Therefore the only ideals (homomorphism kernels) of a field are $$\{0\}$$ and $$F$$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.

More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $$x$$ in a ring $$R$$ to be $$x=ku$$, and the behavior of $$x$$ under any group homomorphism $$g$$ is determined by the behavior of $$k$$ and of $$u$$ under $$g$$, since $$g(x)=g(k)g(u)$$. This explains why $$u$$ is called a "unit" since we can measure behavior of element $$x$$ in terms of what coefficient $$k$$ is in front of $$u$$ when we express $$x$$ in terms of $$u$$.

Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:

Theorem (alternate definition of a unit): $$u\in R$$ is a unit iff for all $$x\in R$$, there exists a $$v\in R$$ such that $$(xv)u=x$$.

The fowards case of this theorem is proven above, while the backward case is provable letting $$x=1$$.